An airline charges the following baggage fees: $25 for the first bag and $35 for the second. Suppose 51% of passengers have no checked luggage, 33% have one piece of checked luggage and 16% have two pieces. We suppose a negligible portion of people check more than two bags. Required:
a. Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.
b. About how much revenue should the airline expect for a flight of 120 passengers? With what standard deviation? Note any assumptions you make and if you think they are justified.

The average revenue per passenger is about $13.85

μ = $13.85

The corresponding standard deviation is $14.51

σ = $14.51

The airline should expect revenue of $1,662 with a standard deviation of $14.51 for a flight of 120 passengers.

Expected revenue = $1,662 ± 14.51

Step-by-step explanation:

An airline charges the following baggage fees:

$25 for the first bag and $35 for the second

Suppose 51% of passengers have no checked luggage,

P(0) = 0.51

33% have one piece of checked luggage and 16% have two pieces.

P(1) = 0.33

P(2) = 0.16

a. Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.

The average revenue per passenger is given by

μ = 0×P(0) + 25×P(1) + 35×P(2)

μ = 0×0.51 + 25×0.33 + 35×0.16

μ = 0 + 8.25 + 5.6

μ = $13.85

Therefore, the average revenue per passenger is about $13.85

The corresponding standard deviation is given by

σ = √σ²

Where σ² is the variance and is given by

σ² = (0 - 13.85)²×0.51 + (25 - 13.85)²×0.33 + (35 - 13.85)²×0.16

σ² = 97.83 + 41.03 + 71.57

σ² = 210.43

So,

σ = √210.43

σ = $14.51

Therefore, the corresponding standard deviation is $14.51

b. About how much revenue should the airline expect for a flight of 120 passengers? With what standard deviation?

For 120 passengers,

Expected revenue = 120×$13.85

Expected revenue = $1,662 ± 14.51

Therefore, the airline should expect revenue of $1,662 with a standard deviation of $14.51 for a flight of 120 passengers.